On average, Tamela has noticed that 20 trains pass by her house daily ( 24 hours) on the nearby train tracks. What is the probability that at most 6 trains will pass her house in a 11 - hour time period? (Round all of your answers to three decimal places.)
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To calculate the probability that at most 6 trains will pass by Tamela's house in an 11-hour period, we first need to determine the average number of trains that pass in that timeframe. Given that 20 trains pass in 24 hours, the average rate is: \[ \lambda = \frac{20 \text{ trains}}{24 \text{ hours}} \times 11 \text{ hours} \approx 9.17 \text{ trains}. \] We can model the number of trains passing as a Poisson distribution \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \) where \( \lambda \) is the average rate and \( k \) is the number of occurrences. Thus, we want to find: \[ P(X \leq 6) = \sum_{k=0}^{6} P(X=k) = \sum_{k=0}^{6} \frac{e^{-9.17} (9.17)^k}{k!} \] Calculating these values for \( k = 0 \) to \( k = 6 \): \[ P(X=0) = \frac{e^{-9.17} (9.17)^0}{0!} \approx 0.000115, \] \[ P(X=1) = \frac{e^{-9.17} (9.17)^1}{1!} \approx 0.001055, \] \[ P(X=2) = \frac{e^{-9.17} (9.17)^2}{2!} \approx 0.004813, \] \[ P(X=3) = \frac{e^{-9.17} (9.17)^3}{3!} \approx 0.014684, \] \[ P(X=4) = \frac{e^{-9.17} (9.17)^4}{4!} \approx 0.036395, \] \[ P(X=5) = \frac{e^{-9.17} (9.17)^5}{5!} \approx 0.070710, \] \[ P(X=6) = \frac{e^{-9.17} (9.17)^6}{6!} \approx 0.115719. \] Now summing these probabilities, we get: \[ P(X \leq 6) \approx 0.000115 + 0.001055 + 0.004813 + 0.014684 + 0.036395 + 0.070710 + 0.115719 \approx 0.243, \] Therefore, the probability that at most 6 trains will pass by in 11 hours is approximately **0.243** when rounded to three decimal places.
